AB is the diameter of the circle with center O. OD is parallel to BC and ∠AOD = 60°. Calculate the numerical values of :

(i) ∠ABD,

(ii) ∠DBC,

(iii) ∠ADC.

Join BD.

(i) From figure,

⇒ ∠BDA = 90° [Angle in semi-circle is a right angle.]

In △OAD,

OA = OD [Radius of same circle]

∠OAD = ∠ODA = x (let) [As angles opposite to equal side are equal]

In △OAD,

⇒ ∠OAD + ∠ODA + ∠AOD = 180° [Angle sum property of triangle]

⇒ x + x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\dfrac{120°}{2}$ = 60°.

From figure,

∠ODB = ∠BDA - ∠ADO = 90° - 60° = 30°.

Given, OD || BC

∠DBC = ∠ODB = 30° [Alternate angles are equal]

Hence, ∠DBC = 30°.

(ii) We know that,

Angle in a semi-circle is a right angle.

∴ ∠BDA = 90°.

Since,

∠OAD = ∠ODA = ∠AOD = 60°.

From figure,

⇒ ∠BDA = ∠ODA + ∠ODB

⇒ 90° = 60° + ∠ODB

⇒ ∠ODB = 90° - 60° = 30°.

Given,

OD || BC

∴ ∠DBC = ∠ODB = 30° [Alternate angles are equal].

Hence, ∠DBC = 30°.

(iii) From figure,

∠ABC = ∠ABD + ∠DBC = 30° + 30° = 60°.

Since, sum of opposite nagles in a cyclic quadrilateral = 180°.

∴ ∠ABC + ∠ADC = 180°

⇒ 60° + ∠ADC = 180°

⇒ ∠ADC = 120°.

Hence, ∠ADC = 120°.